Question

m- u.tourg Lečevim proosa Problem #2 The picture below depicts the motion of my = 3.00 kg prior to its collision with m2 = 8.00 kg. The initial speed of m is 6.00 m/s, and m2 is initially motionless. Both masses have frictionless surfaces, and the coefficient of restitution between mi and ma is 0.70. (a) Write down equations which express conservation of momentum along the x- and y- directions, also indicated in the figure. (b) Express the final velocities of mi and m2 as vectors in ij form. 40 m2

Answer 1

- we can see from the question that line of impact will be as given in diagram below given me =3kg 1812 6413 mza 8 kg V Sinyo I V Cos40 v cesto my l mi ? Line of impact a) momentum conservation along an direction, Line of impact is along y-direction, so no momentum will be transferred along x direction, so, momentum of me will not change along x-direction 80, (before collision) - (After collision), directions Now, momentum conservation along y-direction. Egg musinuo (-ſ) to = mvi + mat (2) b) Cofficient of restitution velocity of saparation along - line of impact - Velocity of approach along line " of impact

0.70 esim Tv sin wo)ch) sa - TV = 269 19) — (8) solving equation (2) - 11:57(-5) = 377" to -(4) from equation (3) and (4) 377 - 3v1 = -8.07 h 30, +8u = -1.67 subtracting , then slav = -3.5 À ✓ = -40.3) i 0.21 (3) Now, so = 0.81 (-7) - 2.6915) - So, velocity of mass me after collision ini = vcasto (4) + v? w = 4.59 () -3(0)

velocity of mass my after collision T = o(%) + 2 The 0.31 (-5) this is velocity of mass me.